## University Calculus: Early Transcendentals (3rd Edition)

$$y'=\frac{1}{3}\Big(\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}\Big)^{1/3}\Big(\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-2}-\frac{2x}{x^2+1}-\frac{2}{2x+3}\Big)$$
$$y=\sqrt[3]{\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}}=\Big(\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}\Big)^{1/3}$$ We have $$y'=\Big(\Big(\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}\Big)^{1/3}\Big)'=\frac{1}{3}\Big(\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}\Big)^{-2/3}\Big(\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}\Big)'$$ $$y'=\frac{1}{3}\Big(\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}\Big)^{-2/3}\times (N)'$$ Consider $N$: $$N=\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}$$ 1) Take the natural logarithm of both sides and simplify: $$\ln N=\ln\Big(\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}\Big)$$ Recall that $\ln(A/B)=\ln A-\ln B$ and $\ln(A\times B)=\ln A+\ln B$: $$\ln N=\ln x(x+1)(x-2)-\ln (x^2+1)(2x+3)$$ $$\ln N =\ln x+\ln(x+1)+\ln(x-2)-\ln(x^2+1)-\ln(2x+3)$$ 2) Then we take the derivative of both sides with respect to $x$ and remember that $(\ln x)'=1/x$: $$\frac{1}{N}\times (N)'=\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-2}-\frac{1}{x^2+1}(x^2+1)'-\frac{1}{2x+3}(2x+3)'$$ $$\frac{N'}{N}=\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-2}-\frac{2x}{x^2+1}-\frac{2}{2x+3}$$ 3) Solve for $N'$: $$N'=\Big(\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-2}-\frac{2x}{x^2+1}-\frac{2}{2x+3}\Big)\times N$$ 4) Substitute for $N=\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}$ $$N'=\Big(\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-2}-\frac{2x}{x^2+1}-\frac{2}{2x+3}\Big)\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}$$ 5) Finally, substitute $N'$ back to $y'$: $$y'=\frac{1}{3}\Big(\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}\Big)^{-2/3}\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}\Big(\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-2}-\frac{2x}{x^2+1}-\frac{2}{2x+3}\Big)$$ $$y'=\frac{1}{3}\Big(\frac{x(x+1)(x-2)}{(x^2+1)(2x+3)}\Big)^{1/3}\Big(\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-2}-\frac{2x}{x^2+1}-\frac{2}{2x+3}\Big)$$